3.16.90 \(\int -\frac {24 e^{\frac {1}{5} (e^{32}-2 e^{32} \log (2)+e^{32} \log ^2(2))}}{64+16 e^{\frac {1}{5} (e^{32}-2 e^{32} \log (2)+e^{32} \log ^2(2))} x+e^{\frac {2}{5} (e^{32}-2 e^{32} \log (2)+e^{32} \log ^2(2))} x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {3}{1+\frac {1}{8} e^{\frac {1}{5} e^{32} (1-\log (2))^2} x} \]

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Rubi [A]  time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.57, number of steps used = 4, number of rules used = 3, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 27, 32} \begin {gather*} \frac {3\ 2^{3+\frac {2 e^{32}}{5}}}{x e^{\frac {1}{5} e^{32} \left (1+\log ^2(2)\right )}+2^{3+\frac {2 e^{32}}{5}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24*E^((E^32 - 2*E^32*Log[2] + E^32*Log[2]^2)/5))/(64 + 16*E^((E^32 - 2*E^32*Log[2] + E^32*Log[2]^2)/5)*x
+ E^((2*(E^32 - 2*E^32*Log[2] + E^32*Log[2]^2))/5)*x^2),x]

[Out]

(3*2^(3 + (2*E^32)/5))/(2^(3 + (2*E^32)/5) + E^((E^32*(1 + Log[2]^2))/5)*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (3\ 2^{3-\frac {2 e^{32}}{5}} e^{\frac {1}{5} e^{32} \left (1+\log ^2(2)\right )}\right ) \int \frac {1}{64+16 e^{\frac {1}{5} \left (e^{32}-2 e^{32} \log (2)+e^{32} \log ^2(2)\right )} x+e^{\frac {2}{5} \left (e^{32}-2 e^{32} \log (2)+e^{32} \log ^2(2)\right )} x^2} \, dx\right )\\ &=-\left (\left (3\ 2^{3-\frac {2 e^{32}}{5}} e^{\frac {1}{5} e^{32} \left (1+\log ^2(2)\right )}\right ) \int \frac {2^{\frac {4 e^{32}}{5}}}{\left (2^{3+\frac {2 e^{32}}{5}}+e^{\frac {e^{32}}{5}+\frac {1}{5} e^{32} \log ^2(2)} x\right )^2} \, dx\right )\\ &=-\left (\left (3\ 2^{3+\frac {2 e^{32}}{5}} e^{\frac {1}{5} e^{32} \left (1+\log ^2(2)\right )}\right ) \int \frac {1}{\left (2^{3+\frac {2 e^{32}}{5}}+e^{\frac {e^{32}}{5}+\frac {1}{5} e^{32} \log ^2(2)} x\right )^2} \, dx\right )\\ &=\frac {3\ 2^{3+\frac {2 e^{32}}{5}}}{2^{3+\frac {2 e^{32}}{5}}+e^{\frac {1}{5} e^{32} \left (1+\log ^2(2)\right )} x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 44, normalized size = 1.57 \begin {gather*} \frac {3\ 2^{3+\frac {2 e^{32}}{5}}}{2^{3+\frac {2 e^{32}}{5}}+e^{\frac {1}{5} e^{32} \left (1+\log ^2(2)\right )} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24*E^((E^32 - 2*E^32*Log[2] + E^32*Log[2]^2)/5))/(64 + 16*E^((E^32 - 2*E^32*Log[2] + E^32*Log[2]^2
)/5)*x + E^((2*(E^32 - 2*E^32*Log[2] + E^32*Log[2]^2))/5)*x^2),x]

[Out]

(3*2^(3 + (2*E^32)/5))/(2^(3 + (2*E^32)/5) + E^((E^32*(1 + Log[2]^2))/5)*x)

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fricas [A]  time = 0.75, size = 28, normalized size = 1.00 \begin {gather*} \frac {24}{x e^{\left (\frac {1}{5} \, e^{32} \log \relax (2)^{2} - \frac {2}{5} \, e^{32} \log \relax (2) + \frac {1}{5} \, e^{32}\right )} + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-24*exp(1/5*exp(16)^2*log(2)^2-2/5*exp(16)^2*log(2)+1/5*exp(16)^2)/(x^2*exp(1/5*exp(16)^2*log(2)^2-2
/5*exp(16)^2*log(2)+1/5*exp(16)^2)^2+16*x*exp(1/5*exp(16)^2*log(2)^2-2/5*exp(16)^2*log(2)+1/5*exp(16)^2)+64),x
, algorithm="fricas")

[Out]

24/(x*e^(1/5*e^32*log(2)^2 - 2/5*e^32*log(2) + 1/5*e^32) + 8)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-24*exp(1/5*exp(16)^2*log(2)^2-2/5*exp(16)^2*log(2)+1/5*exp(16)^2)/(x^2*exp(1/5*exp(16)^2*log(2)^2-2
/5*exp(16)^2*log(2)+1/5*exp(16)^2)^2+16*x*exp(1/5*exp(16)^2*log(2)^2-2/5*exp(16)^2*log(2)+1/5*exp(16)^2)+64),x
, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -24*exp(exp(32)*ln(2)^2/5-exp(32)*ln(2)*
2/5+exp(32)/5)*2*1/16/sqrt(-exp(1/5*(exp(32)*ln(2)^2-2*exp(32)*ln(2)+exp(32)))^2+exp(1/5*(2*exp(32)*ln(2)^2-4*
exp(32)*ln(2)+2*exp(3

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maple [A]  time = 0.37, size = 35, normalized size = 1.25

method	result	size
gosper	\(\frac {24}{x \,{\mathrm e}^{\frac {{\mathrm e}^{32} \ln \relax (2)^{2}}{5}-\frac {2 \,{\mathrm e}^{32} \ln \relax (2)}{5}+\frac {{\mathrm e}^{32}}{5}}+8}\)	\(35\)
risch	\(\frac {24 \,2^{\frac {2 \,{\mathrm e}^{32}}{5}} {\mathrm e}^{-\frac {{\mathrm e}^{32} \left (\ln \relax (2)^{2}+1\right )}{5}}}{8 \,2^{\frac {2 \,{\mathrm e}^{32}}{5}} {\mathrm e}^{-\frac {{\mathrm e}^{32} \left (\ln \relax (2)^{2}+1\right )}{5}}+x}\)	\(43\)
meijerg	\(-\frac {3 \,{\mathrm e}^{\frac {{\mathrm e}^{32} \ln \relax (2)^{2}}{5}-\frac {2 \,{\mathrm e}^{32} \ln \relax (2)}{5}+\frac {{\mathrm e}^{32}}{5}} x}{8 \left (1+2^{-3-\frac {2 \,{\mathrm e}^{32}}{5}} x \,{\mathrm e}^{\frac {{\mathrm e}^{32} \ln \relax (2)^{2}}{5}+\frac {{\mathrm e}^{32}}{5}}\right )}\)	\(52\)
norman	\(-\frac {3 \,{\mathrm e}^{\frac {{\mathrm e}^{32} \ln \relax (2)^{2}}{5}} 2^{-\frac {2 \,{\mathrm e}^{32}}{5}} {\mathrm e}^{\frac {{\mathrm e}^{32}}{5}} x}{x \,{\mathrm e}^{\frac {{\mathrm e}^{32} \ln \relax (2)^{2}}{5}-\frac {2 \,{\mathrm e}^{32} \ln \relax (2)}{5}+\frac {{\mathrm e}^{32}}{5}}+8}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-24*exp(1/5*exp(16)^2*ln(2)^2-2/5*exp(16)^2*ln(2)+1/5*exp(16)^2)/(x^2*exp(1/5*exp(16)^2*ln(2)^2-2/5*exp(16
)^2*ln(2)+1/5*exp(16)^2)^2+16*x*exp(1/5*exp(16)^2*ln(2)^2-2/5*exp(16)^2*ln(2)+1/5*exp(16)^2)+64),x,method=_RET
URNVERBOSE)

[Out]

24/(x*exp(1/5*exp(16)^2*ln(2)^2-2/5*exp(16)^2*ln(2)+1/5*exp(16)^2)+8)

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maxima [B]  time = 0.53, size = 63, normalized size = 2.25 \begin {gather*} \frac {24 \, e^{\left (\frac {1}{5} \, e^{32} \log \relax (2)^{2} + \frac {2}{5} \, e^{32} \log \relax (2) + \frac {1}{5} \, e^{32}\right )}}{x e^{\left (\frac {2}{5} \, e^{32} \log \relax (2)^{2} + \frac {2}{5} \, e^{32}\right )} + 8 \, e^{\left (\frac {1}{5} \, e^{32} \log \relax (2)^{2} + \frac {2}{5} \, e^{32} \log \relax (2) + \frac {1}{5} \, e^{32}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-24*exp(1/5*exp(16)^2*log(2)^2-2/5*exp(16)^2*log(2)+1/5*exp(16)^2)/(x^2*exp(1/5*exp(16)^2*log(2)^2-2
/5*exp(16)^2*log(2)+1/5*exp(16)^2)^2+16*x*exp(1/5*exp(16)^2*log(2)^2-2/5*exp(16)^2*log(2)+1/5*exp(16)^2)+64),x
, algorithm="maxima")

[Out]

24*e^(1/5*e^32*log(2)^2 + 2/5*e^32*log(2) + 1/5*e^32)/(x*e^(2/5*e^32*log(2)^2 + 2/5*e^32) + 8*e^(1/5*e^32*log(
2)^2 + 2/5*e^32*log(2) + 1/5*e^32))

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mupad [B]  time = 1.21, size = 1, normalized size = 0.04 \begin {gather*} \mathrm {NaN} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(24*exp(exp(32)/5 - (2*exp(32)*log(2))/5 + (exp(32)*log(2)^2)/5))/(16*x*exp(exp(32)/5 - (2*exp(32)*log(2)
)/5 + (exp(32)*log(2)^2)/5) + x^2*exp((2*exp(32))/5 - (4*exp(32)*log(2))/5 + (2*exp(32)*log(2)^2)/5) + 64),x)

[Out]

NaN

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sympy [B]  time = 0.23, size = 80, normalized size = 2.86 \begin {gather*} \frac {24 \cdot 2^{\frac {2 e^{32}}{5}} e^{\frac {e^{32} \log {\relax (2 )}^{2}}{5}} e^{\frac {e^{32}}{5}}}{x e^{\frac {2 e^{32} \log {\relax (2 )}^{2}}{5}} e^{\frac {2 e^{32}}{5}} + 8 \cdot 2^{\frac {2 e^{32}}{5}} e^{\frac {e^{32} \log {\relax (2 )}^{2}}{5}} e^{\frac {e^{32}}{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-24*exp(1/5*exp(16)**2*ln(2)**2-2/5*exp(16)**2*ln(2)+1/5*exp(16)**2)/(x**2*exp(1/5*exp(16)**2*ln(2)*
*2-2/5*exp(16)**2*ln(2)+1/5*exp(16)**2)**2+16*x*exp(1/5*exp(16)**2*ln(2)**2-2/5*exp(16)**2*ln(2)+1/5*exp(16)**
2)+64),x)

[Out]

24*2**(2*exp(32)/5)*exp(exp(32)*log(2)**2/5)*exp(exp(32)/5)/(x*exp(2*exp(32)*log(2)**2/5)*exp(2*exp(32)/5) + 8
*2**(2*exp(32)/5)*exp(exp(32)*log(2)**2/5)*exp(exp(32)/5))

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